help

\(\frac{1}{12}{y}^{2}-3\)

r + r cosθ = 6

r = √[x^2 + y^2]

cos θ =  x/r 

So we have that

√[x^2 + y^2]  +  r (x/r)  =  6

√[x^2 + y^2] + x  = 6

√[x^2 + y^2]  =  6 - x        square both sides

x^2  + y^2  = x^2 - 12x + 36

y^2  = -12x + 36

12x = -y^2 + 36

x = -y^2/12 + 3

Thanks Chris, I was wondering how to do that.